Integral and its limit Evaluate: $$ \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$
Rewriting this as $$I(n)= \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ and then Differentiating under the Integral Sign with respect to $n$. I was unable to go much further with this.
$$$$ Original problem: $$ \large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$
 A: Expand $\cos^{\frac1n}x$ in powers of $\frac{1}{n}$ using that
$$\cos^{\varepsilon}x=1+\varepsilon\ln\cos x+O\left(\varepsilon^2\right).$$
Now the limit becomes equal to
$$\lim_{n\to \infty}n\int_{0}^{\pi/2}\left(1-\cos^{\frac1n}x\right)dx=-\int_0^{\pi/2}\ln\cos x\,dx=\frac{\pi\ln 2}{2}.$$
P.S. The integral can also be evaluated in terms of gamma functions (see here), but this is of little help for evaluating the limit.
A: I think you should solve the actual problem by solving the limit first.
The limit can be taken into the integral, so now the problem becomes
$$\int_{0}^{\pi /2}\lim_{n\rightarrow \infty }n(1-\sqrt[n]{\cos x})dx$$
By L'Hôpital's rule,
$$\lim_{n \to \infty }n(1-\sqrt[n]{\cos x})=\lim_{n \to \infty }\frac{-\cos^{\frac 1n}x\cdot \ln \cos x\cdot \frac {-1}{n^2}}{-\frac{1}{n^2}}=-\ln \cos x$$
Hence we have 
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=-\int_{0}^{\pi /2}\ln \cos x dx$$
Now the integral on the right side is $-\frac{\pi }{2}\ln2$. You can refer to here.
Thus
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=\frac{\pi }{2}\ln2$$
A: $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{\pi/2}\sqrt[n]{\sin x}\,dx = \int_{0}^{1}\frac{u^{1/n}}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt$$
so by using the Beta function we have:
$$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{2\,\Gamma\left(1+\frac{1}{2n}\right)}.\tag{1}$$
Let now $f(z)=\frac{\Gamma\left(\frac{1}{2}+z\right)}{\Gamma\left(1+z\right)}$. By the properties of the digamma function we have:
$$\frac{f'(z)}{f(z)}=\frac{d}{dz}\log f(z) = H_{x-\frac{1}{2}}-H_{x}\tag{2}$$
hence:
$$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \frac{\sqrt{\pi}}{2}\left(\sqrt{\pi}-\frac{\sqrt{\pi}\log 2}{n}+O\left(\frac{1}{n^2}\right)\right)\tag{3}$$
and it follows that:
$$ \lim_{n\to +\infty} n\int_{0}^{\pi/2}\left(1-\sqrt[n]{\cos x}\right)\,dx =\color{red}{\frac{\pi}{2}\log 2.}\tag{4}$$
